3.432 \(\int \sec ^4(c+d x) (a+b \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=46 \[ \frac {(a+b) \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {b \tan ^5(c+d x)}{5 d} \]

[Out]

a*tan(d*x+c)/d+1/3*(a+b)*tan(d*x+c)^3/d+1/5*b*tan(d*x+c)^5/d

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Rubi [A]  time = 0.04, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3675, 373} \[ \frac {(a+b) \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {b \tan ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + b*Tan[c + d*x]^2),x]

[Out]

(a*Tan[c + d*x])/d + ((a + b)*Tan[c + d*x]^3)/(3*d) + (b*Tan[c + d*x]^5)/(5*d)

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \sec ^4(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \left (1+x^2\right ) \left (a+b x^2\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (a+(a+b) x^2+b x^4\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {a \tan (c+d x)}{d}+\frac {(a+b) \tan ^3(c+d x)}{3 d}+\frac {b \tan ^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 53, normalized size = 1.15 \[ \frac {\tan (c+d x) \left (5 a \tan ^2(c+d x)+15 a+3 b \sec ^4(c+d x)-b \sec ^2(c+d x)-2 b\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + b*Tan[c + d*x]^2),x]

[Out]

(Tan[c + d*x]*(15*a - 2*b - b*Sec[c + d*x]^2 + 3*b*Sec[c + d*x]^4 + 5*a*Tan[c + d*x]^2))/(15*d)

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fricas [A]  time = 0.49, size = 56, normalized size = 1.22 \[ \frac {{\left (2 \, {\left (5 \, a - b\right )} \cos \left (d x + c\right )^{4} + {\left (5 \, a - b\right )} \cos \left (d x + c\right )^{2} + 3 \, b\right )} \sin \left (d x + c\right )}{15 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

1/15*(2*(5*a - b)*cos(d*x + c)^4 + (5*a - b)*cos(d*x + c)^2 + 3*b)*sin(d*x + c)/(d*cos(d*x + c)^5)

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giac [A]  time = 1.52, size = 48, normalized size = 1.04 \[ \frac {3 \, b \tan \left (d x + c\right )^{5} + 5 \, a \tan \left (d x + c\right )^{3} + 5 \, b \tan \left (d x + c\right )^{3} + 15 \, a \tan \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c)^2),x, algorithm="giac")

[Out]

1/15*(3*b*tan(d*x + c)^5 + 5*a*tan(d*x + c)^3 + 5*b*tan(d*x + c)^3 + 15*a*tan(d*x + c))/d

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maple [A]  time = 0.59, size = 66, normalized size = 1.43 \[ \frac {b \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )-a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+b*tan(d*x+c)^2),x)

[Out]

1/d*(b*(1/5*sin(d*x+c)^3/cos(d*x+c)^5+2/15*sin(d*x+c)^3/cos(d*x+c)^3)-a*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c))

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maxima [A]  time = 0.31, size = 39, normalized size = 0.85 \[ \frac {3 \, b \tan \left (d x + c\right )^{5} + 5 \, {\left (a + b\right )} \tan \left (d x + c\right )^{3} + 15 \, a \tan \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+b*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

1/15*(3*b*tan(d*x + c)^5 + 5*(a + b)*tan(d*x + c)^3 + 15*a*tan(d*x + c))/d

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mupad [B]  time = 11.94, size = 40, normalized size = 0.87 \[ \frac {\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}+\left (\frac {a}{3}+\frac {b}{3}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+a\,\mathrm {tan}\left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x)^2)/cos(c + d*x)^4,x)

[Out]

(tan(c + d*x)^3*(a/3 + b/3) + a*tan(c + d*x) + (b*tan(c + d*x)^5)/5)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+b*tan(d*x+c)**2),x)

[Out]

Integral((a + b*tan(c + d*x)**2)*sec(c + d*x)**4, x)

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